Question

If the partial fraction of (2x + 1)/(x ^ 2 * (x + 1)) are A/(x + 1) + 1/x + 1/(x ^ 2) then the value of A,​

Answer 1

Answer:

1/[(x-1)(x+1)]= a/(x-1) + b/(x+1)

1/[(x-1)(x+1)]= [a(x+1)+b(x-1)]/[(x-1)(x+1)]

Cancelling denominator

1=a(x+1)+b(x-1)

1=ax + a +bx -b

1=(a+b)x + (a-b)

Equating on both sides

a+b=0

And a-b=1

Solving we get

2a=1

a=1/2

Put value of a in a+b=0

Then we get 1/2 + b=0

b=-1/2

So partial fractions is

1/[(x-1)(x+1)]= (1/2)/(x-1) - (1/2)/(x+1)

Step-by-step explanation:

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Answer 2

Answer:

If the partial fraction of (2x + 1)/(x ^ 2 * (x + 1)) are A/(x + 1) + 1/x + 1/(x ^ 2) then the value of A,